### 3.444 $$\int \frac{(d+e x)^m}{(b x+c x^2)^2} \, dx$$

Optimal. Leaf size=180 $-\frac{c^2 (d+e x)^{m+1} (2 c d-b e (2-m)) \, _2F_1\left (1,m+1;m+2;\frac{c (d+e x)}{c d-b e}\right )}{b^3 (m+1) (c d-b e)^2}+\frac{(d+e x)^{m+1} (2 c d-b e m) \, _2F_1\left (1,m+1;m+2;\frac{e x}{d}+1\right )}{b^3 d^2 (m+1)}-\frac{(d+e x)^{m+1} (c x (2 c d-b e)+b (c d-b e))}{b^2 d \left (b x+c x^2\right ) (c d-b e)}$

[Out]

-(((d + e*x)^(1 + m)*(b*(c*d - b*e) + c*(2*c*d - b*e)*x))/(b^2*d*(c*d - b*e)*(b*x + c*x^2))) - (c^2*(2*c*d - b
*e*(2 - m))*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (c*(d + e*x))/(c*d - b*e)])/(b^3*(c*d - b*e)^
2*(1 + m)) + ((2*c*d - b*e*m)*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, 1 + (e*x)/d])/(b^3*d^2*(1 +
m))

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Rubi [A]  time = 0.18127, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.21, Rules used = {740, 830, 65, 68} $-\frac{c^2 (d+e x)^{m+1} (2 c d-b e (2-m)) \, _2F_1\left (1,m+1;m+2;\frac{c (d+e x)}{c d-b e}\right )}{b^3 (m+1) (c d-b e)^2}+\frac{(d+e x)^{m+1} (2 c d-b e m) \, _2F_1\left (1,m+1;m+2;\frac{e x}{d}+1\right )}{b^3 d^2 (m+1)}-\frac{(d+e x)^{m+1} (c x (2 c d-b e)+b (c d-b e))}{b^2 d \left (b x+c x^2\right ) (c d-b e)}$

Antiderivative was successfully veriﬁed.

[In]

Int[(d + e*x)^m/(b*x + c*x^2)^2,x]

[Out]

-(((d + e*x)^(1 + m)*(b*(c*d - b*e) + c*(2*c*d - b*e)*x))/(b^2*d*(c*d - b*e)*(b*x + c*x^2))) - (c^2*(2*c*d - b
*e*(2 - m))*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (c*(d + e*x))/(c*d - b*e)])/(b^3*(c*d - b*e)^
2*(1 + m)) + ((2*c*d - b*e*m)*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, 1 + (e*x)/d])/(b^3*d^2*(1 +
m))

Rule 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b
*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 830

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m, (f + g*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !RationalQ[m]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{(d+e x)^m}{\left (b x+c x^2\right )^2} \, dx &=-\frac{(d+e x)^{1+m} (b (c d-b e)+c (2 c d-b e) x)}{b^2 d (c d-b e) \left (b x+c x^2\right )}-\frac{\int \frac{(d+e x)^m ((c d-b e) (2 c d-b e m)-c e (2 c d-b e) m x)}{b x+c x^2} \, dx}{b^2 d (c d-b e)}\\ &=-\frac{(d+e x)^{1+m} (b (c d-b e)+c (2 c d-b e) x)}{b^2 d (c d-b e) \left (b x+c x^2\right )}-\frac{\int \left (\frac{(-c d+b e) (-2 c d+b e m) (d+e x)^m}{b x}+\frac{c^2 d (-2 c d+b e (2-m)) (d+e x)^m}{b (b+c x)}\right ) \, dx}{b^2 d (c d-b e)}\\ &=-\frac{(d+e x)^{1+m} (b (c d-b e)+c (2 c d-b e) x)}{b^2 d (c d-b e) \left (b x+c x^2\right )}+\frac{\left (c^2 (2 c d-b e (2-m))\right ) \int \frac{(d+e x)^m}{b+c x} \, dx}{b^3 (c d-b e)}-\frac{(2 c d-b e m) \int \frac{(d+e x)^m}{x} \, dx}{b^3 d}\\ &=-\frac{(d+e x)^{1+m} (b (c d-b e)+c (2 c d-b e) x)}{b^2 d (c d-b e) \left (b x+c x^2\right )}-\frac{c^2 (2 c d-b e (2-m)) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac{c (d+e x)}{c d-b e}\right )}{b^3 (c d-b e)^2 (1+m)}+\frac{(2 c d-b e m) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;1+\frac{e x}{d}\right )}{b^3 d^2 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.147617, size = 174, normalized size = 0.97 $-\frac{(d+e x)^{m+1} \left (b^2 d (m+1) (c d-b e)^2+x (b+c x) \left (c^2 d^2 (b e (m-2)+2 c d) \, _2F_1\left (1,m+1;m+2;\frac{c (d+e x)}{c d-b e}\right )-(c d-b e)^2 (2 c d-b e m) \, _2F_1\left (1,m+1;m+2;\frac{e x}{d}+1\right )\right )+b c d (m+1) x (b e-2 c d) (b e-c d)\right )}{b^3 d^2 (m+1) x (b+c x) (c d-b e)^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(d + e*x)^m/(b*x + c*x^2)^2,x]

[Out]

-(((d + e*x)^(1 + m)*(b^2*d*(c*d - b*e)^2*(1 + m) + b*c*d*(-2*c*d + b*e)*(-(c*d) + b*e)*(1 + m)*x + x*(b + c*x
)*(c^2*d^2*(2*c*d + b*e*(-2 + m))*Hypergeometric2F1[1, 1 + m, 2 + m, (c*(d + e*x))/(c*d - b*e)] - (c*d - b*e)^
2*(2*c*d - b*e*m)*Hypergeometric2F1[1, 1 + m, 2 + m, 1 + (e*x)/d])))/(b^3*d^2*(c*d - b*e)^2*(1 + m)*x*(b + c*x
)))

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Maple [F]  time = 0.61, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( ex+d \right ) ^{m}}{ \left ( c{x}^{2}+bx \right ) ^{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^m/(c*x^2+b*x)^2,x)

[Out]

int((e*x+d)^m/(c*x^2+b*x)^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x\right )}^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(c*x^2+b*x)^2,x, algorithm="maxima")

[Out]

integrate((e*x + d)^m/(c*x^2 + b*x)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (e x + d\right )}^{m}}{c^{2} x^{4} + 2 \, b c x^{3} + b^{2} x^{2}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(c*x^2+b*x)^2,x, algorithm="fricas")

[Out]

integral((e*x + d)^m/(c^2*x^4 + 2*b*c*x^3 + b^2*x^2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m/(c*x**2+b*x)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x\right )}^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(c*x^2+b*x)^2,x, algorithm="giac")

[Out]

integrate((e*x + d)^m/(c*x^2 + b*x)^2, x)